Question

Given that the ka for hocl is 3.5 × 10–8, calculate the k value for the reaction of hocl with oh–.

Answer 1

Following reaction is involved in above system
HOCl(aq)  ↔  H+(aq) + OCl-(aq)
OCl-(aq) + H2O(l)  ↔ HOCl(aq) + OH-(aq)

Now, if the system is obeys 1st order kinetics we have
K  = [OCl-][H+]/[HOCl]   ............. (1)
∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8)   ............. (2)

and now considering that system is obeying 2nd order kinetics, we have
K  = [OH-][HOCl-] / [OCl-] ................. (3)
Subs 2 in 3 we get
K = [OH-][H+] (1 / 3.0 * 10-8)
we know that, [OH-][H+] = 10-14

∴K = 3.3 * 10-7

Thus, correct answer is e i.e none of these

Answer 2

Answer:

D

3.5 * 10^6

Explanation:

In this question, we are asked to calculate the the k-value for the reaction of HOCl with OH- given Ka value for HOCl

Firstly, we write the dissociation equation for HOCl

This is as follows;

HOCl ——> H+ + OCl-

Now, we write an expression for Ka

Ka = [H+][OCl-]/HOCl = 3.5 * 10^-8

Now, we write an equation for the reaction of HOCl and OH- ;

HOCl + OH- —-> OCl- + H2O(l)

The K value expression for this can be written as;

K = [OCl-]/[HOCl][OH-]

= [H+][OCl-]/[HOCl] * [1/[H+][OH-]

= ka * 1/Kw = 3.5 * 10^-8 /(1 * 10^-14) = 3.5 * 10^6